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In an A.P. of $40$ terms, the sum of first $9$ terms is $153$ and the sum of last $6$ terms is $687$. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.
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Sol. Here $n = 40$,
$S_9 = \frac{9}{2} [2a +8d] = 153 \Rightarrow a + 4d = 17$ ----- (i)
and $S_{40} - S_{34} = 687$ or $a_{35} + a_{36} + a_{37} + a_{38} + a_{39} + a_{40} = 687$
$\Rightarrow 6a+219d = 687$ or $2a + 73d = 229$ ----- (ii)
solving (i) and (ii) to get $a = 5, d = 3$
Also, $S_{40} = \frac{40}{2} (10 + 39 \times 3) = 2540$
$S_9 = \frac{9}{2} [2a +8d] = 153 \Rightarrow a + 4d = 17$ ----- (i)
and $S_{40} - S_{34} = 687$ or $a_{35} + a_{36} + a_{37} + a_{38} + a_{39} + a_{40} = 687$
$\Rightarrow 6a+219d = 687$ or $2a + 73d = 229$ ----- (ii)
solving (i) and (ii) to get $a = 5, d = 3$
Also, $S_{40} = \frac{40}{2} (10 + 39 \times 3) = 2540$